gas volume conversion.pdf

7/27/2019 gas volume conversion.pdf

http://slidepdf.com/reader/full/gas-volume-conversionpdf 1/9

CHE 205 – Chemical Process Principles

Section 3: EPCP, Chapter 5

3-1Copyright Richard M. Felder, Lisa G. Bullard, and Michael D. Dickey (2009)

EQUATIONS OF STATE FOR GASES

Questions

A gas enters a reactor at a rate of 255 SCMH. What does that mean?

An orifice meter mounted in a process gas line indicates a flow rate of 24 ft

3

/min. The gastemperature is 195oF and the pressure is 62 psig. The gas is a mixture containing 70 mole%Co and the balance H2. What is the mass flow rate of the hydrogen in the gas?

A reactor feed stream consists of O2 flowing at 32 kg/s. The gas is to be compressed from37oC and 2.8 atm absolute to 54oC and 284 atm. What are the volumetric flow rates at theinlet and outlet (needed to rate the compressor)?

A pitot tube indicates that the velocity of a stack gas is 5.0 m/s at 175oC. The stack diameter is 4.0 m. A continuous stack analyzer indicates an SO2 level of 2500 ppm (2500 molesSO2/106 moles gas). At what rate in kg/s is SO2 being discharged into the atmosphere?

A 70.0 m3 tank is rated at 2000 kPa. If 150 kg of helium is charged into the tank, what willthe pressure be? How much more helium can be added before the rated pressure is attained?

Answers: Need an equation of state: relationship between temperature (T ), pressure (P), volume (V ), and number of moles (n) of a gas.

In Chapter 4, streams on flow charts labeled like this:

In this chapter, stream data just as likely to look like this:

For material balances, however, we still need moles and mole fractions. The job now becomes one of converting volumetric flow rates (or volumes) to molar flow rates (or moles), and (for gases) partial pressures to mole fractions. (Latter is easy: y A = p A/P )

Convert volumes to moles

– Solids & liquids: use tabulated densities (volume to mass) & molecular weights (mass to moles).Mixtures—either look up mixture density data or assume volume additivity & calculate densityfrom Eq. (5.1-1).

– Gases, can’t use tabulated densities. (Why not?) Instead, need an equation of state (EOS) — aformula relating V , n, T , and P. Simplest is the ideal gas EOS.

Read Section 5.1. We won’t lecture on it, but you need to know it.

Ideal Gas Equation of State (Sections 5.2a & 5.2b)

(batch) or (continuous)

ˆ ˆor where / (or / ) is the of the gas

PV nRT PV nRT

PV RT V V n V n specific molar volume

100 mol/s

0.600 mol A/mol0.400 mol B/mol

250 L/s @ 37o

C, 800 mm Hg

p A = 420 mm Hg (partial pressure of A)





3-2

Copyright Richard M. Felder, Lisa G. Bullard, and Michael D. Dickey (2009)

T and P must be absolute temperature (K, oR) and absolute pressure (not gauge). R is the gas

constant —values given on inside back cover of text.

Convenient —applies regardless of what the gas is, & whether the gas has a single component or is amixture. If you can’t assume ideal gas behavior (i.e., if gas is nonideal or real),

Approximate —greatest validity at low gas densities (high T , low P), when gas molecules are far enough apart for intermolecular forces to be negligible (behave like billiard balls). Usually ok for temperatures at or above 0oC & pressures at or below 1 atm. Rule of thumb for when to use it given in

Eqs. (5.2-3) on p. 192. Three will get you four. Given any three of the variables , (or ), , and (or )P V V T n n , calculate the

fourth one.

Example

The volumetric flow rate of a stream of propane at 150oC and 70.0 atm being fed to a combustion furnace

is measured and found to be 29.0 m3/h.

(a) Determine its molar flow rate.

Solution. We are given three of the four gas law variables (_____, ______, and ______) and so candetermine the fourth one (________).

From the inside back cover, 0.08206 (L atm)/(mol K) R .

3

3 870.0 atm 29.0 m mol K kmol C H

h 0.08206 L atmh

kmol= 58.5

h

PV n

RT

(b) Now, suppose an analysis of the combustion chamber products shows that the molar flow rate of the propane was 101 kmol/h. Think of four possible reasons for the discrepancy between the two stated values of the molar flow rate.

1. _______________________________________________________________________

2. _______________________________________________________________________

3. _______________________________________________________________________

4. _______________________________________________________________________

(c) Check the assumption of ideal gas behavior.

Solution:





3-3


0.08206 L atm L __________ , therefore from Eq. ____________,

mol K mol

ideal gas beh

RT

P

avior is a ___________ assumption.

For another example, work through Problem 5.10 in the workbook.

Standard temperature and pressure (STP): 0oC (273.16K, 491.67oR), 1 atm. At STP, 1 mol occupies22.415 L, 1 kmol occupies 22.415 m3, & 1 lb-mole occupies 359.05 ft3. (Memorize)

Suppose you are told that a gas flows at a rate of, say, 1280 SCFH [standard cubic feet per hour, or ft3(STP)/h] (a) It does not mean that the gas is at standard temperature and pressure. It does mean that if you

brought it from whatever its temperature and pressure really are to 0oC and 1 atm, its volumetricflow rate would be 1280 ft3/h. (See Example 5.2-4.)

(b) You can calculate the molar flow rate of the gas as

3

3

1280 ft (STP) 1 lb-mole lb-mole3.56

hh 359.05 ft (STP)n

(c) If you know that the actual temperature of the gas is 120oC (393K) and 0.800 atm, you cancalculate its actual flow rate as

3 31280 ft (STP) 393 K 1 atm ft2300

h 273 K 0.800 atm hV

(See Example 5.2-3.)





3-4


Ideal Gas Mixtures, Partial Pressures, and Volume Percentages (Section 5.2c)

Suppose yA = mole fraction of a component A in a mixture of gases at pressure P and volume V .(Example: A mixture of gases at P = 1000 mm Hg with a volume of 200 liters contains 30 mole% CH4,

50 mole% C2H6, and 20 mole% C2H4.)

Partial pressure of a component of a gas:

pA = yAP Example:

4CH 0.30 1000 mm Hg = 300 mm Hg p

The partial pressures of all components of a mixture add up to the total pressure (prove it).

Pure component volume and percentage by volume: The pure component volume of A is thevolume A would occupy if it were by itself at the mixture temperature and pressure

Divide A A A A A

Pv n RT v n y

PV nRT V n

(volume fraction = mole fraction)

The percentage by volume (% v/v) is 100 times the volume fraction. Thus,

% v/v = mole% for an ideal gas mixture

% v/v has no practical significance for a nonideal gas mixture

Three alternative ways of telling you the value of a mole fraction

From now on, any of these specifications may be given in material balance problems. You should immediately convert the first two to mole fractions when you label the flow chart, and if you label the

partial pressure on the flow chart also label the mole fraction and count pA = yAP as another equation inthe degree-of-freedom analysis.

MOLEFRACTIONS yA

PARTIALPRESSURE

pA = yAP

VOLUME

FRACTION

% v/v = mole%for an ideal gas

mixture





3-5


Exercise. Liquid acetone (C3H6O) is fed at a rate of 400 L/min into a heated chamber where it evaporatesinto a nitrogen stream which enters at 27oC and 475 mm Hg (gauge). The gas leaving the heater is diluted by another nitrogen stream flowing at a measured rate of 419 m3(STP) at 25oC and 2.5 atm. Thecombined gases are then compressed to a total pressure P = 6.3 atm (gauge) at a temperature of 325oC.The partial pressure of acetone in this stream is pa = 501 torr (501 mm Hg). Atmospheric pressure is 763

torr.

(a) Write the complete set of equations you would solve to determine the molar composition of thestream product gas stream and the volumetric flow rate of the nitrogen entering the evaporator.

(b) How is it possible for the second nitrogen stream to be at standard temperature and pressure and at 25oC and 2.5 atm?

Solution. Verify that the flow chart is completely labeled.

(a) Degree-of-freedom analysis:

System equations:

(b) _________________________________________________________________________________

3

1

1 2

o

(m /min)

(mol N / min)

27 C, 475 torr

V

n

400 L/min C3H6O (l)

2 (mol/min)n

Evaporator

419 m3(STP) N2/min

3 2(mol N / min)n

25oC, 2.5 atm

Compressor 4 (mol/min)n

y4 (mol C3H6O(v)/mol)

(1– y4) (mol N2/mol)325oC, 6.3 atm (gauge)

pa = 501 torr





3-6


Nonideal (Real) Gases (Section 5.3)

When you can’t assume ideal gas behavior (low T , high P, outside criteria of Eqs. 5.2-3), need to use amore complex equation of state, all of which incorporate properties of the species in the gas. (Ideal gasEOS is independent of species.) Read p. 5-4 of the workbook.

Note definitions of critical temperature and pressure in Section 5.3a. You can look up T c and Pc inTable B.1.

Virial equations of state (Section 5.3b). Mostly of theoretical interest—rarely used in practice.

Cubic equations of state and the SRK equation (Section 5.3c). Commonly used for single species (for mixtures, use compressibility factor EOS). The SRK equation of state (Eq. 5.3-7) may be the mostcommonly used EOS other than the ideal gas equation. Easy or hard to use, depending on which of

the three variables ˆ( , , )P V T is unknown.

Procedure:

For given species, look up T c , Pc , and the Pitzer acentric factor (Table 5.3-1 for selected

species). Calculate a, b, and m from Eqs. 5.3-8, 5.3-9, & 5.3-10.

If T and V̂ are known, evaluate T r from Eq. 5.3-11 and from Eq. 5.3-12, solve Eq. 5.3-7 for P.

If T and P are known, enter Eq. 5.3-7 in E-Z Solve, enter all known values, and solve for V̂ .

(Alternatively, use Goal Seek in Excel, as in Example 5.3-3.)

If P and V̂ are known, enter Eqs. 5.3-7, 5.3-11 for T r , and 5.3-12 for into E-Z Solve, enter all

known values, and solve for T.

Compressibility factor equation of state (Section 5.4): PV = znRT , where z is the compressibility

factor (a fudge factor — the farther it is from 1, the farther the gas is from ideal).

Either you have tables where you can look up z for a given T and P (Section 5.4a, single speciesonly), or you use the law of corresponding states (Section 5.4b) to estimate z.

Procedure: Given two of the variables T , P, and V̂ (or V and n or and V n )

1. Look up T c and Pc (e.g. in Table B.1). Apply Newton’s corrections for H2, He (p. 208).

2. Calculate two of the quantities reduced temperature. T r = T/T c , reduced pressure Pr = P/Pc , and

ideal reduced volume,ideal

ˆˆ

/r

c c

V V

RT P , depending on which two of the variables T , P, and V̂ are

known.

3. Look up z on one of the generalized compressibility charts, Figs. 5.4-1 – 5.4-4.

4. Substitute known variables and z into the compressibility factor equation of state to determine theunknown variable.





3-7


Example: Example 5.4-2, p. 209.

Note: If the gas were anything other than nitrogen, then Z will change because the criticalconstants are different. For example, Z = 1.4 for hydrogen.

Go through Test Yourself on p. 210.

Kay’s rule: PVT calculations for nonideal gas mixtures using compressibility charts (Section 5.4c):Calculate pseudocritical temperature and pseudocritical pressure by weighting T ci and Pci by molefractions of it h component (Eqs. 5.4-9 and 5.4-10), then proceed as for single component. This is the

only method we will present in this book for doing PVT calculations on nonideal gas mixtures.





3-8


Exercise. A natural gas (85 mole% CH4, 15% C2H6) at 20oC and 80 atm is burned completely with 30%

excess air. natural gas 285 L/s.V The stack gas emerges at 280oC and 1 atm. What is stack gas ?V

Solution.

(a) Draw and label the flow chart.

(b) What equations of state should you use to relate the volumetric flow rates of the fuel and stack gasesto their molar flow rates?

Fuel gas: __________________________________________________

Stack gas: _________________________________________________

(c) Do the degree-of-freedom analysis (use atomic balances).

Furnace

CH4 + 2O2 CO2 + 2H2O

C2H6 + 72

O2 2CO2 + 3H2O





3-9


(d) Write out the seven equations required to solve for the unknowns, letting (T cm , Pcm) and (T ce , Pce) =critical temperatures of methane and ethane, respectively. (You could find their values in Table B1 butdon’t bother.)

gas volume conversion.pdf

Documents